Two old RSM students were standing ten feet apart when they ran in opposite directions with speeds of 180 feet per minute and 16

Question

2 years ago

7

0 feet per minute respectively, how many minutes would pass before they were 1870 feet apart?

1 answer:

inessss [21] · Answer 1 · 2023-04-13T01:09:45+03:00

The number of minutes that would pass before they were 1870 feet apart is 5.47 minutes

Since one student runs at a speed of 180 feet per minutes, in time t minutes, he moves a distance of d = 180t.
Also, the second student runs at a speed of 160 feet per minute, in time t minutes, he moves a distance of d' = 160t.

Since they are initially 10 feet apart, their total distance apart after t minutes is D = d + 10 + d'

D = 180t + 10 + 160t

D = 340t + 10

<h3>Number of minutes before they are 1870 feet</h3>

Making t subject of the formula, we have

t = (D - 10)/340

Since they are 1870 feet apart after t minutes, D = 1870 feet.

t = (D - 10)/340

t = (1870 - 10)/340

t = 1860/340

t = 5.47 minutes

So, the number of minutes that would pass before they were 1870 feet apart is 5.47 minutes